Sharp power-type Heronian mean bounds for the Sándor and Yang means

Abstract

We prove that the double inequalities $H_{\alpha}(a, b) < X(a, b)<H_{\beta}(a, b)$ and $H_{\lambda}(a, b)< U(a, b)<H_{\mu}(a, b)$ hold for all $a, b>0$ with $a\neq b$ if and only if $\alpha\leq1/2$, $\beta\geq\log3/(1+\log2)=0.6488\cdots$, $\lambda\leq2\log3/(2\log\pi-\log2) =1.3764\cdots$, and $\mu\geq2$, where $H_{p}(a, b)$, $X(a, b)$, and $U(a, b)$ are, respectively, the pth power-type Heronian mean, Sándor mean, and Yang mean of a and b.

1 Introduction

For $p\in\mathbb{R}$, the pth power-type Heronian mean $H_{p}(a, b)$ of two positive real numbers a and b is defined by

$$ H_{p}(a, b)= \biggl[\frac{a^{p}+(ab)^{p/2}+b^{p}}{3} \biggr]^{1/p} \quad(p\neq 0),\qquad H_{0}(a, b)=\sqrt{ab}. $$

(1.1)

It is well known that $H_{p}(a, b)$ is continuous and strictly increasing with respect to $p\in\mathbb{R}$ for fixed $a, b>0$ with $a\neq b$.

Let $G(a, b)=\sqrt{ab}$, $L(a, b)=(a-b)/(\log a-\log b)$, $P(a, b)=(a-b)/[2\arcsin((a-b)/(a+b))]$, $I(a, b)=(a^{a}/b^{b})^{1/(a-b)}/e$, $A(a, b)=(a+b)/2$, $T(a, b)=(a-b)/[2\arctan((a-b)/(a+b))]$, $Q(a, b)=\sqrt{(a^{2}+b^{2})/2}$ and $M_{r}(a, b)=[(a^{r}+b^{r})/2]^{1/r}$ ($r\neq0$), and $M_{0}(a, b)=\sqrt{ab}$ be, respectively, the geometric, logarithmic, first Seiffert, identric, arithmetic, second Seiffert, quadratic, and rth power means of two distinct positive real numbers a and b. Then it is well known that the inequalities

$$\begin{aligned} G(a,b)&=M_{0}(a,b)< L(a,b)<P(a,b)<I(a,b) \\ &<A(a,b)=M_{1}(a,b)<T(a,b)<Q(a,b)=M_{2}(a,b) \end{aligned}$$

hold for all $a, b>0$ with $a\neq b$.

Let $a, b>0$. Then the Sándor mean $X(a,b)$ [1] and Yang mean $U(a,b)$ [2] are given by

$$ X(a,b)=A(a,b)e^{\frac{G(a,b)}{P(a,b)}-1} $$

(1.2)

and

$$ U(a,b)=\frac{a-b}{\sqrt{2}\arctan (\frac{a-b}{\sqrt{2ab}} )} \quad(a\neq b),\qquad U(a, a)=a, $$

(1.3)

respectively.

The Yang mean $U(a,b)$ is the special case of the Seiffert type mean $T_{M, p}(a, b)=(a-b)/[p\arctan((a-b)/(pM(a,b)))]$ defined by Toader in [3], where $M(a, b)$ is a bivariate mean and p is a positive real number. Indeed, $U(a,b)=T_{G,\sqrt{2}}(a,b)$. Recently, the power-type Heronian, Sándor, and Yang means have been the subject of intensive research.

For all $a, b>0$ with $a\neq b$, Yang [4] and Sándor [5] proved that the double inequality

$$ M_{1/2}(a,b)< H_{1}(a,b)<I(a,b) $$

holds, and the inequality $H_{1}(a,b)< M_{2/3}(a,b)$ can be found in the literature [6].

Jia and Cao [7] proved that the inequalities

$$\begin{aligned} &L(a,b)< H_{p}(a,b)<M_{q}(a,b), \\ &A(a,b)=M_{1}(a,b)<H_{\log3/\log2}(a,b) \end{aligned}$$

(1.4)

hold for all $a, b>0$ with $a\neq b$ if $p\geq1/2$ and $q\geq2p/3$. Inequality (1.4) can also be found in the literature [8], p.64 and [9].

In [10], the authors proved that the double inequality

$$ H_{p}(a,b)< T(a,b)<H_{q}(a,b) $$

holds for all $a, b>0$ with $a\neq b$ if and only if $p\leq\log3/(\log \pi-\log2)$ and $q\geq5/2$.

Sándor [11] presented the inequalities

$$\begin{aligned}& X(a,b)< \frac{P^{2}(a,b)}{A(a,b)}, \qquad\frac {A(a,b)G(a,b)}{P(a,b)}<X(a,b)<\frac{A(a,b)P(a,b)}{2P(a,b)-G(a,b)}, \\& X(a,b)>\frac{A(a,b)L(a,b)}{P(a,b)}e^{\frac{G(a,b)}{L(a,b)}-1},\qquad X(a,b)>\frac{A(a,b)[P(a,b)+G(a,b)]}{3P(a,b)-G(a,b)}, \\& \frac{A^{2}(a,b)G(a,b)}{P(a,b)L(a,b)}e^{\frac {L(a,b)}{A(a,b)}-1}<X(a,b)<A(a,b) \biggl[\frac{1}{e}+ \biggl(1-\frac {1}{e} \biggr)\frac{G(a,b)}{P(a,b)} \biggr], \\& A(a,b)+G(a,b)-P(a,b)<X(a,b)<A^{-1/3}(a,b) \biggl[\frac {A(a,b)+G(a,b)}{2} \biggr]^{4/3}, \\& P^{1/(\log\pi-\log2)}(a,b)A^{1-1/(\log\pi-\log 2)}(a,b)<X(a,b)<P^{-1}(a,b) \biggl[ \frac{A(a,b)+G(a,b)}{2} \biggr]^{2} \end{aligned}$$

for all $a, b>0$ with $a\neq b$.

Yang et al. [12] proved that the double inequality

$$ M_{p}(a,b)< X(a,b)<M_{q}(a,b) $$

holds for all $a, b>0$ with $a\neq b$ if and only if $p\leq1/3$ and $q\geq\log2/(1+\log2)$.

In [2], Yang established the inequalities

$$\begin{aligned}& P(a,b)< U(a,b)<T(a,b), \qquad \frac{G(a,b)T(a,b)}{A(a,b)}<U(a,b)<\frac{P(a,b)Q(a,b)}{A(a,b)}, \\& Q^{1/2}(a,b) \biggl[\frac{2G(a,b)+Q(a,b)}{3} \biggr]^{1/2}<U(a,b)<Q^{2/3}(a,b) \biggl[\frac{G(a,b)+Q(a,b)}{2} \biggr]^{1/3}, \\& \frac{G(a,b)+Q(a,b)}{2}<U(a,b)< \biggl[\frac{2}{3} \biggl(\frac {G(a,b)+Q(a,b)}{2} \biggr)^{1/2}+\frac{1}{3}Q^{1/2}(a,b) \biggr]^{2} \end{aligned}$$

for all $a, b>0$ with $a\neq b$.

In [13, 14], the authors proved that the double inequalities

$$\begin{aligned} & \biggl[\frac{2}{3} \biggl(\frac{G(a,b)+Q(a,b)}{2} \biggr)^{p}+ \frac {1}{3}Q^{p}(a,b) \biggr]^{1/p} \\ &\quad< U(a,b)< \biggl[\frac{2}{3} \biggl(\frac{G(a,b)+Q(a,b)}{2} \biggr)^{q}+\frac {1}{3}Q^{q}(a,b) \biggr]^{1/q}, \\ &\frac{2^{1-\lambda}(G(a,b)+Q(a,b))^{\lambda}Q(a,b)+G(a,b)Q^{\lambda }(a,b)}{2^{1-\lambda}(G(a,b)+Q(a,b))^{\lambda}+Q^{\lambda}(a,b)} \\ &\quad<U(a,b)<\frac{2^{1-\mu}(G(a,b)+Q(a,b))^{\mu}Q(a,b)+G(a,b)Q^{\mu }(a,b)}{2^{1-\mu}(G(a,b)+Q(a,b))^{\mu}+Q^{\mu}(a,b)}, \\ &M_{\alpha}(a,b)<U(a,b)<M_{\beta}(a,b) \end{aligned}$$

hold for all $a, b>0$ with $a\neq b$ if and only if $p\leq p_{0}$, $q\geq1/5$, $\lambda\geq1/5$, $\mu\leq p_{1}$, $\alpha\leq2\log 2/(2\log\pi-\log2)$, and $\beta\geq4/3$, where $p_{0}=0.1941\cdots$ is the unique solution of the equation $p\log(2/\pi)-\log (1+2^{1-p} )+\log3=0$ on the interval $(1/10, \infty)$, and $p_{1}=\log(\pi-2)/\log2=0.1910\cdots$.

The main purpose of this paper is to present the best possible parameter α, β, λ, and μ such that the double inequalities $H_{\alpha}(a, b)< X(a, b)<H_{\beta}(a, b)$ and $H_{\lambda}(a, b)< U(a, b)<H_{\mu}(a, b)$ hold for all $a, b>0$ with $a\neq b$.

2 Lemmas

In order to prove our main results we need two lemmas, which we present in this section.

Lemma 2.1

Let $p\in(0, 1)$ and

$$\begin{aligned} f(x)={}&(p-1)x^{3p+2}+(4p-3)x^{2p+2}+(p-3)x^{p+2}+6x^{4p}+6x^{3p}-6x^{p} \\ &{}+2x^{4p-2}+(3-p)x^{3p-2}+(3-4p)x^{2p-2}+(1-p)x^{p-2}-2x^{2}-6. \end{aligned}$$

(2.1)

Then the following statements are true:

• (1) if $p=1/2$, then $f(x)<0$ for all $x>1$;

• (2) if $p=\log3/(1+\log2)=0.6488\cdots$, then there exists $\lambda \in(1, \infty)$ such that $f(x)>0$ for $x\in(1, \lambda)$ and $f(x)<0$ for $x\in(\lambda, \infty)$.

Proof

For part (1), if $p=1/2$, then (2.1) becomes

$$ f(x)=-\frac{(x-1)(\sqrt {x}-1)^{2}}{2x^{3/2}} \bigl(x^{3}+4x^{5/2}+13x^{2}+16x^{3/2}+13x+4 \sqrt{x}+1 \bigr). $$

(2.2)

Therefore, part (1) follows from (2.2).

For part (2), let $p=\log3/(1+\log2)$, $f_{1}(x)=f^{\prime}(x)/x$, $f_{2}(x)=x^{5-p}f_{1}^{\prime}(x)$, $f_{3}(x)=f_{2}^{\prime }(x)/(2px)$, and $f_{4}(x)=f_{3}^{\prime}(x)/(2x)$. Then elaborated computations lead to

$$\begin{aligned}& f(1)=0, \qquad\lim_{x\rightarrow+\infty}f(x)=-\infty, \end{aligned}$$

(2.3)

$$\begin{aligned}& \begin{aligned}[b] f_{1}(x)={}&(p-1) (3p+2)x^{3p}+2(p+1) (4p-3)x^{2p}+(p-3) (p+2)x^{p} \\ &{}+24px^{4p-2}+18px^{3p-2}-6px^{p-2}+4(2p-1)x^{4p-4}+(3-p) (3p-2)x^{3p-4} \\ &{}+2(p-1) (3-4p)x^{2p-4}+(1-p) (p-2)x^{p-4}-4, \end{aligned} \\& f_{1}(1)=36(2p-1)>0,\qquad \lim_{x\rightarrow+\infty}f_{1}(x)=- \infty, \end{aligned}$$

(2.4)

$$\begin{aligned}& \begin{aligned}[b] f_{2}(x)={}&3p(p-1) (3p+2)x^{2p+4}+4p(p+1) (4p-3)x^{p+4}+48p(2p-1)x^{3p+2} \\ &{}+18p(3p-2)x^{2p+2}+16(2p-1) (p-1)x^{3p}+(3-p) (3p-2) (3p-4)x^{2p} \\ &{}+4(3-4p) (p-1) (p-2)x^{p}+p(p-3) (p+2)x^{4} \\ &{}-6p(p-2)x^{2}+(1-p) (p-2) (p-4), \end{aligned} \\& f_{2}(1)=72(2p-1)^{2}>0,\qquad \lim_{x\rightarrow+\infty}f_{2}(x)=- \infty, \end{aligned}$$

(2.5)

$$\begin{aligned}& \begin{aligned}[b] f_{3}(x)={}&3(p-1) (p+2) (3p+2)x^{2p+2}+2(p+1) (p+4) (4p-3)x^{p+2} \\ &{}+24(2p-1) (3p+2)x^{3p}+18(p+1) (3p-2)x^{2p}+24(2p-1) (p-1)x^{3p-2} \\ &{}+(3-p) (3p-2) (3p-4)x^{2p-2}+2(p-1) (p-2) (3-4p)x^{p-2} \\ &{}+2(p-3) (p+2)x^{2}-6(p-2), \end{aligned} \\& f_{3}(1)=12 \bigl(31p^{2}-12p-5 \bigr)>0, \qquad\lim _{x\rightarrow+\infty }f_{3}(x)=-\infty, \end{aligned}$$

(2.6)

$$\begin{aligned}& \begin{aligned}[b] f_{4}(x)={}&3(p-1) (p+1) (p+2) (3p+2)x^{2p}+(p+1) (p+2) (p+4) (4p-3)x^{p} \\ &{}+36p(2p-1) (3p+2)x^{3p-2}+18p(p+1) (3p-2)x^{2p-2} \\ &{}+12(p-1) (2p-1) (3p-2)x^{3p-4} \\ &{}+(p-1) (3-p) (3p-2) (3p-4)x^{2p-4}+(p-1) (p-2)^{2}(3-4p)x^{p-4} \\ &{}+2(p-3) (p+2). \end{aligned} \end{aligned}$$

(2.7)

It follows from (2.7) and $p=\log3/(1+\log2)=0.6488\cdots$ together with $13p^{4}+337p^{3}-80p^{2}-72=-11.3153\cdots<0$ that

$$\begin{aligned} f_{4}(x)< {}&3(p-1) (p+1) (p+2) (3p+2)+(p+1) (p+2) (p+4) (4p-3) \\ &{}+36p(2p-1) (3p+2)+18p(p+1) (3p-2)x^{2p-2}+12(p-1) (2p-1) (3p-2) \\ &{}+(p-1) (3-p) (3p-2) (3p-4)x^{2p-4}+(p-1) (p-2)^{2}(3-4p)x^{p-4} \\ &{}+2(p-3) (p+2) \\ ={}&18p(p+1) (3p-2)x^{2p-2}+(p-1) (3-p) (3p-2) (3p-4)x^{2p-4} \\ &{}+(p-1) (p-2)^{2}(3-4p)x^{p-4}+ \bigl(13p^{4}+337p^{3}-80p^{2}-72 \bigr)<0 \end{aligned}$$

(2.8)

for $x\in(1, \infty)$.

Inequality (2.8) implies that $f_{3}(x)$ is strictly decreasing on $[1, \infty)$. Then (2.6) leads to the conclusion that there exists $\lambda _{1}\in(1, \infty)$ such that $f_{2}(x)$ is strictly increasing on $[1, \lambda_{1}]$ and strictly decreasing on $[\lambda_{1}, \infty)$.

It follows from the piecewise monotonicity of $f_{2}$ and (2.5) that there exists $\lambda_{2}\in(\lambda_{1}, \infty)$ such that $f_{1}(x)$ is strictly increasing on $[1, \lambda_{2}]$ and strictly decreasing on $[\lambda_{2}, \infty)$.

From (2.4) and the piecewise monotonicity of $f_{1}$ we clearly see that there exists $\lambda_{3}\in(\lambda_{2}, \infty)$ such that $f(x)$ is strictly increasing on $[1, \lambda_{3}]$ and strictly decreasing on $[\lambda_{3}, \infty)$.

Therefore, part (2) follows from (2.3) and the piecewise monotonicity of f. □

Lemma 2.2

Let $p\in(1, 2]$ and

$$\begin{aligned}& \begin{aligned}[b] g(x)={}&2x^{4p+4}-2x^{4p+2}+10x^{4p}+6x^{4p-2}+(p-1)x^{3p+6}-2x^{3p+4}-2x^{3p+2} \\ &{}+14x^{3p}+(7-p)x^{3p-2}+4(p-1)x^{2p+6}-12x^{2p+4}+12x^{2p}+4(1-p)x^{2p-2} \\ &{}+(p-7)x^{p+6}-14x^{p+4}+2x^{p+2}+2x^{p}+(1-p)x^{p-2}-2 \bigl(3x^{6}+5x^{4}-x^{2}+1 \bigr). \end{aligned} \end{aligned}$$

(2.9)

Then the following statements are true:

• (1) if $p=2$, then $g(x)>0$ for all $x>1$;

• (2) if $p=2\log3/(2\log\pi-\log2)=1.3764\cdots$, then there exists $\mu\in(1, \infty)$ such that $g(x)<0$ for $x\in(1, \mu)$ and $g(x)>0$ for $x\in(\mu, \infty)$.

Proof

For part (1), if $p=2$, then (2.9) becomes

$$ g(x)=3 \bigl(x^{4}-1 \bigr)^{3}. $$

(2.10)

Therefore, part (1) follows from (2.10).

For part (2), let $p=2\log3/(2\log\pi-\log2)$, $g_{1}(x)=g^{\prime }(x)/x$, $g_{2}(x)=g_{1}^{\prime}(x)/x$, $g_{3}(x)=g_{2}^{\prime }(x)/x$, and $g_{4}(x)=x^{9-p}g_{3}^{\prime}(x)$. Then elaborated computations lead to

$$\begin{aligned} &g(1)=0,\qquad \lim_{x\rightarrow\infty}g(x)=+\infty, \end{aligned}$$

(2.11)

$$\begin{aligned} &g_{1}(x)=8(p+1)x^{4p+2}-4(2p+1)x^{4p}+40px^{4p-2}+12(2p-1)x^{4p-4}+3(p-1) (p+2)x^{3p+4} \\ &\hphantom{g_{1}(x)=}{}-2(3p+4)x^{3p+2}-2(3p+2)x^{3p}+42px^{3p-2}+(7-p) (3p-2)x^{3p-4} \\ &\hphantom{g_{1}(x)=}{}+8(p-1) (p+3)x^{2p+4} \\ &\hphantom{g_{1}(x)=}{}-24(p+2)x^{2p+2}+24px^{2p-2}-8(1-p)^{2}x^{2p-4}+(p-7) (p+6)x^{p+4} \\ &\hphantom{g_{1}(x)=}{}-14(p+4)x^{p+2}+2(p+2)x^{p}+2px^{p-2}+(1-p) (p-2)x^{p-4}-36x^{4}-40x^{2}+4, \\ &g_{1}(1)=-144(2-p)< 0, \qquad\lim_{x\rightarrow\infty}g_{1}(x)=+ \infty, \end{aligned}$$

(2.12)

$$\begin{aligned} &g_{2}(x)=16(p+1) (2p+1)x^{4p}-16p(2p+1)x^{4p-2} +80p(2p-1)x^{4p-4} \\ &\hphantom{g_{2}(x)=}{}+48(p-1)(2p-1)x^{4p-6} \\ &\hphantom{g_{2}(x)=}{}+3(p-1) (p+2) (3p+4)x^{3p+2}-2(3p+2) (3p+4)x^{3p}-6p(3p+2)x^{3p-2} \\ &\hphantom{g_{2}(x)=}{}+42p(3p-2)x^{3p-4} +(7-p) (3p-2) (3p-4)x^{3p-6} \\ &\hphantom{g_{2}(x)=}{}+16(p-1) (p+2) (p+3)x^{2p+2}-48(p+1) (p+2)x^{2p}+48p(p-1)x^{2p-4} \\ &\hphantom{g_{2}(x)=}{}-16(1-p)^{2}(p-2)x^{2p-6}+(p-7) (p+4) (p+6)x^{p+2}-14(p+2) (p+4)x^{p} \\ &\hphantom{g_{2}(x)=}{}+2p(p+2)x^{p-2} +2p(p-2)x^{p-4}+(1-p) (p-2) (p-4)x^{p-6}-144x^{2}-80, \\ &g_{2}(1)=-288 \bigl(2+3p-2p^{2} \bigr)< 0,\qquad \lim _{x\rightarrow\infty }g_{2}(x)=+\infty, \end{aligned}$$

(2.13)

$$\begin{aligned} & g_{3}(x)=64p(p+1) (2p+1)x^{4p-2}-32p \bigl(4p^{2}-1 \bigr)x^{4p-4}+320p(p-1) (2p-1)x^{4p-6} \\ &\hphantom{g_{3}(x)=}{}+96(p-1) (2p-1) (2p-3)x^{4p-8}+3(p-1) (p+2) (3p+2) (3p+4)x^{3p} \\ &\hphantom{g_{3}(x)=}{}-6p(3p+2) (3p+4)x^{3p-2} -6p \bigl(9p^{2}-4 \bigr)x^{3p-4}+42p(3p-2) (3p-4)x^{3p-6} \\ &\hphantom{g_{3}(x)=}{}+3(7-p) (p-2) (3p-2) (3p-4)x^{3p-8} \\ &\hphantom{g_{3}(x)=}{}+32 \bigl(p^{2}-1 \bigr) (p+2) (p+3)x^{2p}-96p(p+1) (p+2)x^{2p-2}+96p(p-1) (p-2)x^{2p-6} \\ &\hphantom{g_{3}(x)=}{}-32(1-p)^{2}(p-2) (p-3)x^{2p-8}+(p-7) (p+2) (p+4) (p+6)x^{p} \\ &\hphantom{g_{3}(x)=}{}-14p(p+2) (p+4)x^{p-2} +2p \bigl(p^{2}-4 \bigr)x^{p-4}+2p(p-2) (p-4)x^{p-6} \\ &\hphantom{g_{3}(x)=}{}+(1-p) (p-2) (p-4) (p-6)x^{p-8}-288, \\ &g_{3}(1)=48 \bigl(43p^{3}-100p^{2}+58p-36 \bigr)< 0, \qquad\lim_{x\rightarrow\infty }g_{3}(x)=+ \infty, \end{aligned}$$

(2.14)

$$\begin{aligned} &g_{4}(x)= 128p(p+1) \bigl(4p^{2}-1 \bigr)x^{3p+6}-128p \bigl(4p^{2}-1 \bigr) (p-1)x^{3p+4} \\ &\hphantom{g_{4}(x)=}{}+640p(p-1) (2p-1) (2p-3)x^{3p+2}+384(p-1) (p-2) (2p-1) (2p-3)x^{3p} \\ &\hphantom{g_{4}(x)=}{}+9p(p-1) (p+2) (3p+2) (3p+4)x^{2p+8}-6p \bigl(9p^{2}-4 \bigr) (3p+4)x^{2p+6} \\ &\hphantom{g_{4}(x)=}{}-6p \bigl(9p^{2}-4 \bigr) (3p-4)x^{2p+4}+126p(p-2) (3p-2) (3p-4)x^{2p+2} \\ &\hphantom{g_{4}(x)=}{}+3(7-p) (p-2) (3p-2) (3p-4) (3p-8)x^{2p}+64p \bigl(p^{2}-1 \bigr) (p+2) (p+3)x^{p+8} \\ &\hphantom{g_{4}(x)=}{}-192p \bigl(p^{2}-1 \bigr) (p+2)x^{p+6}+192p(p-1) (p-2) (p-3)x^{p+2} \\ &\hphantom{g_{4}(x)=}{}-64(1-p)^{2}(p-2) (p-3) (p-4)x^{p} \\ &\hphantom{g_{4}(x)=}{}+p(p-7) (p+2) (p+4) (p+6)x^{8}-14p \bigl(p^{2}-4 \bigr) (p+4)x^{6}+2p \bigl(p^{2}-4 \bigr) (p-4)x^{4} \\ &\hphantom{g_{4}(x)=}{}+2p(p-2) (p-4) (p-6)x^{2}+(1-p) (p-2) (p-4) (p-6) (p-8) \\ &\hphantom{g_{4}(x)}=:a_{1}x^{3p+6}+a_{4}x^{3p+4} +a_{8}x^{3p+2}+a_{11}x^{3p}+a_{0}x^{2p+8}+a_{3} x^{2p+6}+a_{7}x^{2p+4}+a_{10}x^{2p+2} \\ &\hphantom{g_{4}(x)=}{}+a_{14}x^{2p}+a_{2}x^{p+8}+a_{6}x^{p+6}+a_{13}x^{p+2} +a_{16}x^{p}+a_{5}x^{8}+a_{9}x^{6} \\ &\hphantom{g_{4}(x)=}{}+a_{12}x^{4}+a_{15}x^{2}+a_{17}, \end{aligned}$$

(2.15)

$$\begin{aligned} &\sum_{n=0}^{8}a_{n}=2p \bigl(73p^{4}+1306p^{3}-3344p^{2}+3272p-1328 \bigr)>0, \end{aligned}$$

(2.16)

$$\begin{aligned} &a_{9}+a_{10}=16p \bigl(70p^{3}-287p^{2}+350p-112 \bigr)>0, \end{aligned}$$

(2.17)

$$\begin{aligned} &\sum_{n=11}^{15}a_{n}=-81p^{5}+2839p^{4} -13904p^{3}+25652p^{2}-19600p+4992>0, \end{aligned}$$

(2.18)

$$\begin{aligned} &a_{16}+a_{17}=5 \bigl(-13p^{5}+145p^{4}-608p^{3}+1196p^{2}-1104p+384 \bigr)>0. \end{aligned}$$

(2.19)

Note that

$$ \begin{aligned}[b] 2p+8&>3p+6>p+8>2p+6>3p+4>8>p+6>2p+4\\ &>3p+2>6>2p+2>3p>4>p+2>2p>2>p>1. \end{aligned} $$

(2.20)

It follows from (2.15)-(2.20) that

$$ g_{4}(x)> \Biggl(\sum_{n=0}^{8}a_{n} \Biggr)x^{p+8}+(a_{9}+a_{10})x^{2p+2}+ \Biggl(\sum_{n=11}^{15}a_{n} \Biggr)x^{2p}+(a_{16}+a_{17})x^{p}>0 $$

(2.21)

for $x\in(1, \infty)$.

Therefore, part (2) follows easily from (2.11)-(2.14) and (2.21). □

3 Main results

Theorem 3.1

The double inequality

$$ H_{\alpha}(a, b)< X(a, b)<H_{\beta}(a, b) $$

holds for all $a, b>0$ with $a\neq b$ if and only if $\alpha\leq1/2$ and $\beta\geq\log3/(1+\log2)=0.6488\cdots$.

Proof

Since $X(a,b)$ and $H_{p}(a,b)$ are symmetric and homogeneous of degree one, we assume that $a>b$. Let $x=\sqrt{a/b}\in (1, \infty)$ and $p\in\mathbb{R}$. Then (1.1) and (1.2) lead to

$$\begin{aligned} &\log \bigl[X(a,b) \bigr]-\log \bigl[H_{p}(a,b) \bigr] \\ &\quad=\log \biggl(\frac{x^{2}+1}{2} \biggr)+\frac{2x}{x^{2}-1}\arcsin \biggl( \frac{x^{2}-1}{x^{2}+1} \biggr)-\frac{1}{p}\log \biggl(\frac {x^{2p}+x^{p}+1}{3} \biggr)-1:=F(x). \end{aligned}$$

(3.1)

Simple computations lead to

$$\begin{aligned}& F(1)=0, \end{aligned}$$

(3.2)

$$\begin{aligned}& \lim_{x\rightarrow+\infty}F(x)=\frac{1}{p}\log3-\log2-1, \end{aligned}$$

(3.3)

$$\begin{aligned}& F^{\prime}(x)=\frac{2(1+x^{2})}{(x^{2}-1)^{2}}F_{1}(x), \end{aligned}$$

(3.4)

where

$$\begin{aligned}& F_{1}(x)=\frac {(x^{2}-1)(2x^{2p}+x^{p+2}+x^{p}+2x^{2})}{2x(1+x^{2}) (x^{2p}+x^{p}+1)}-\arcsin \biggl(\frac{1-x^{2}}{1+x^{2}} \biggr), \\& F_{1}(1)=0,\qquad \lim_{x\rightarrow+\infty}F_{1}(x)=+ \infty, \end{aligned}$$

(3.5)

$$\begin{aligned}& F^{\prime}_{1}(x) =-\frac{x^{2}-1}{2(x^{2}+1)^{2}(x^{2p}+x^{p}+1)^{2}}f(x), \end{aligned}$$

(3.6)

where $f(x)$ is defined as in Lemma 2.1.

If $p=1/2$, then from Lemma 2.1(1), (3.1), (3.2), and (3.4)-(3.6) we clearly see that

$$ X(a,b)>H_{1/2}(a,b) $$

for all $a, b>0$ with $a\neq b$.

If $p=\log3/(1+\log2)$, then (3.3) becomes

$$ \lim_{x\rightarrow+\infty}F(x)=0. $$

(3.7)

It follows from Lemma 2.2(2) and (3.6) that there exists $\lambda\in (1, \infty)$ such that $F_{1}(x)$ is strictly decreasing on $[1, \lambda ]$ and strictly increasing on $[\lambda, \infty)$.

Equations (3.4) and (3.5) together with the piecewise monotonicity of $F_{1}$ lead to the conclusion that there exists $\lambda^{\ast}\in(1, \infty)$ such that $F(x)$ is strictly decreasing on $[1, \lambda^{\ast }]$ and strictly increasing on $[\lambda^{\ast}, \infty)$.

Therefore,

$$ X(a,b)< H_{\log3/(1+\log2)}(a,b) $$

for all $a, b>0$ with $a\neq b$ follows from (3.1), (3.2), (3.7), and the piecewise monotonicity of F.

Next, we prove that $\alpha=1/2$ and $\beta=\log3/(1+\log2)$ are the best possible parameters such that the double inequality $H_{\alpha}(a, b)< X(a, b)<H_{\beta}(a, b)$ holds for all $a, b>0$ with $a\neq b$.

If $p<\log3/(1+\log2)$, then (3.3) leads to

$$ \lim_{x\rightarrow+\infty}F(x)>0. $$

(3.8)

Equation (3.1) and inequality (3.8) imply that there exists large enough $T_{0}=T_{0}(p)>1$ such that $X(a,b)>H_{p}(a,b)$ for all $a, b>0$ with $a/b\in(T_{0}, \infty)$ if $p<\log3/(1+\log2)$.

Let $p>1/2$, $x>0$, and $x\rightarrow0$. Then elaborated computations lead to

$$\begin{aligned} &H{_{p}}(1, 1+x)-X(1, 1+x) \\ &\quad= \biggl[\frac{1+(1+x)^{p/2}+(1+x)^{p}}{3} \biggr]^{1/p}- \biggl(1+ \frac{x}{2} \biggr)e^{\frac{2\sqrt{1+x}\arcsin (\frac{x}{2+x} )}{x}-1} \\ &\quad=\frac{2p-1}{24}x^{2}+o \bigl(x^{2} \bigr). \end{aligned}$$

(3.9)

Equation (3.9) implies that there exists small enough $\delta_{0}=\delta _{0}(p)>0$ such that $X(1, 1+x)< H{_{p}}(1, 1+x)$ for $x\in(0, \delta _{0})$ if $p>1/2$. □

Theorem 3.2

The double inequality

$$ H_{\lambda}(a, b)< U(a, b)<H_{\mu}(a, b) $$

holds for all $a, b>0$ with $a\neq b$ if and only if $\lambda\leq2\log 3/(2\log\pi-\log2)=1.3764\cdots$ and $\mu\geq2$.

Proof

Since $U(a,b)$ and $H_{p}(a,b)$ are symmetric and homogeneous of degree one, we assume that $a>b$. Let $x=\sqrt{a/b}\in (1, \infty)$ and $p\in\mathbb{R}$. Then (1.1) and (1.3) lead to

$$\begin{aligned} &\log \bigl[U(a,b) \bigr]-\log \bigl[H_{p}(a,b) \bigr] \\ &\quad=\log \biggl[\frac{x^{2}-1}{\sqrt{2}\arctan (\frac{x^{2}-1}{\sqrt {2}x} )} \biggr]-\frac{1}{p}\log \biggl( \frac{x^{2p}+x^{p}+1}{3} \biggr):=G(x). \end{aligned}$$

(3.10)

Simple computations lead to

$$\begin{aligned}& G(1)=0, \end{aligned}$$

(3.11)

$$\begin{aligned}& \lim_{x\rightarrow+\infty}G(x)=\frac{1}{p}\log3+ \frac{1}{2}\log2-\log \pi, \end{aligned}$$

(3.12)

$$\begin{aligned}& G^{\prime}(x)=\frac {2x^{2p}+x^{p+2}+x^{p}+2x^{2}}{x(x^{2}-1)(x^{2p}+x^{p}+1)\arctan (\frac{x^{2}-1}{\sqrt{2}x} )}G_{1}(x), \end{aligned}$$

(3.13)

where

$$\begin{aligned}& G_{1}(x)=\arctan \biggl(\frac{x^{2}-1}{\sqrt{2}x} \biggr)- \frac{\sqrt {2}x(x^{4}-1)(x^{2p}+x^{p}+1)}{(x^{4}+1)(2x^{2p}+x^{p+2}+x^{p}+2x^{2})}, \\& G_{1}(1)=0,\qquad \lim_{x\rightarrow+\infty}G_{1}(x)=- \infty, \end{aligned}$$

(3.14)

$$\begin{aligned}& G^{\prime}_{1}(x)=-\frac{\sqrt {2}x^{2}(x^{2}-1)}{(x^{4}+1)^{2}(2x^{2p}+x^{p+2}+x^{p}+2x^{2})^{2}}g(x), \end{aligned}$$

(3.15)

where $g(x)$ is defined as in Lemma 2.2.

If $p=2\log3/(2\log\pi-\log2)$, then (3.15) and Lemma 2.2(2) lead to the conclusion that there exists $\mu\in(1, \infty)$ such that $G_{1}(x)$ is strictly increasing on $[1, \mu]$ and strictly decreasing on $[\mu, \infty)$.

It follows from (3.13) and (3.14) together with the piecewise monotonicity of $G_{1}$ that there exists $\mu^{\ast}\in(1, \infty)$ such that $G(x)$ is strictly increasing on $[1, \mu^{\ast}]$ and strictly decreasing on $[\mu^{\ast}, \infty)$.

Note that (3.12) becomes

$$ \lim_{x\rightarrow+\infty}G(x)=0. $$

(3.16)

Therefore,

$$ U(a,b)>H_{2\log3/(2\log\pi-\log2)}(a,b) $$

for all $a, b>0$ with $a\neq b$ follows from (3.10), (3.11), and (3.16) together with the piecewise monotonicity of G.

If $p=2$, then

$$ U(a,b)< H_{2}(a,b) $$

for all $a, b>0$ with $a\neq b$ follows easily from (3.10), (3.11), and (3.13)-(3.15) together with Lemma 2.2(1).

Next, we prove that $\lambda=2\log3/(2\log\pi-\log2)$ and $\mu=2$ are the best possible parameters such that the double inequality

$$ H_{\lambda}(a, b)< U(a, b)<H_{\mu}(a, b) $$

holds for all $a, b>0$ with $a\neq b$.

If $p>2\log3/(2\log\pi-\log2)$, then (3.12) leads to

$$ \lim_{x\rightarrow+\infty}G(x)< 0. $$

(3.17)

Equation (3.10) and inequality (3.17) imply that there exists large enough $T_{1}=T_{1}(p)>1$ such that $U(a, b)< H_{p}(a,b)$ for all $a, b>0$ with $a/b\in(T_{1}, \infty)$.

Let $p<2$, $x>0$, and $x\rightarrow0$. Then elaborated computations lead to

$$\begin{aligned} &U(1, 1+x)-H_{p}(1, 1+x) \\ &\quad=\frac{x}{\sqrt{2}\arctan (\frac{x}{\sqrt{2(1+x)}} )} - \biggl[\frac{1+(1+x)^{p/2}+(1+x)^{p}}{3} \biggr]^{1/p} \\ &\quad=\frac{2-p}{12}x^{2}+o \bigl(x^{2} \bigr). \end{aligned}$$

(3.18)

Inequality (3.18) implies that there exists small enough $\delta _{1}=\delta_{1}(p)>0$ such that $U(1, 1+x)>H_{p}(1, 1+x)$ for $x\in(0, \delta_{1})$. □